Question

How do you find an equation of the tangent line to the curve $y = \dfrac{{{e^x}}}{x}$ at the point $(1,\;e)?$

Answer 1

Firstly find the slope of the required tangent line by differentiating the given curve at the given point. Then use the slope point method to write the equation for the required tangent line. If a line has slope “m” and passing point $(a,\;b)$ then its equation is written as $y - b = m(x - a)$

Complete step by step solution:
In order to find the equation of the tangent line to the given curve $y = \dfrac{{{e^x}}}{x}$ at point $(1,\;e)$, we have to find the slope of the required tangent line and we can find the slope by differentiating the curve at the given point.
So first differentiating the curve $y = \dfrac{{{e^x}}}{x}$ with respect to $x$
$\Rightarrow y = \dfrac{{{e^x}}}{x} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{{{e^x}}}{x}} \right) \\\$
Differentiating this with help of $\dfrac{u}{v}$ rule, we will get
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{{{e^x}}}{x}} \right) \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{xd({e^x})}}{{dx}} - \dfrac{{{e^x}d(x)}}{{dx}}}}{{{x^2}}} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{x{e^x} - {e^x}}}{{{x^2}}} \\\$
Now putting value of $x = 1$ in order to find the slope at point $(1,\;e)$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{x{e^x} - {e^x}}}{{{x^2}}} \\\ \Rightarrow \dfrac{{dy}}{{dx}}\;{\text{at}}\;(1,\;e) = \dfrac{{1 \times {e^x} - {e^x}}}{{{x^2}}} \\\ \Rightarrow \dfrac{{dy}}{{dx}}\;{\text{at}}\;(1,\;e) = \dfrac{0}{{{x^2}}} \\\ \Rightarrow \dfrac{{dy}}{{dx}}\;{\text{at}}\;(1,\;e) = 0 \\\$
So we got the value of slope, $\dfrac{{dy}}{{dx}}\;at\;(1,\;e) = m = 0$
Now writing the required equation of the tangent line of the given curve at the point $(1,\;e)$ with the help of slope $(m)$, passing point $(1,\;e)$ and the slope point method for writing equation of line Slope point method is given as
$y - b = m(x - a)\;{\text{where}}\;m$ is the slope of the line and $(a,\;b)$ is the passing point.
$\therefore$ required equation of tangent line will be given as
$\Rightarrow y - e = 0 \times (x - 1) \\\ \Rightarrow y - e = 0 \\\ \Rightarrow y = e \\\$
Therefore $y = e$ is the required equation of tangent line of the curve $y = \dfrac{{{e^x}}}{x}$ at point $(1,\;e)$

Note: Slope of a given curve at a particular point also gives the direction of that curve at that point.
The $\dfrac{u}{v}$ rule of differentiation is given as follows
$\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}$